Velocity and acceleration algorithm

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SUMMARY

The discussion focuses on the accuracy of velocity and acceleration equations used in numerical simulations, specifically comparing equations (1) and (3) with (2) and (4). The equations (2) and (4) utilize the finite difference method to provide more accurate estimates by reducing time shift errors. The equation (4) incorporates a factor of 7*dt to enhance the smoothing of acceleration calculations, derived from the Verlet algorithm. The participants explore the rationale behind the coefficients in these equations and their implications for computational physics.

PREREQUISITES
  • Understanding of finite difference methods
  • Familiarity with the Verlet algorithm
  • Basic knowledge of kinematics equations
  • Concept of time step (dt) in numerical simulations
NEXT STEPS
  • Research the application of the Verlet algorithm in physics simulations
  • Learn about the derivation and application of finite difference methods
  • Explore the impact of time step size (dt) on numerical accuracy
  • Investigate alternative methods for calculating derivatives in numerical analysis
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Students and professionals in computational physics, numerical analysts, and anyone involved in developing algorithms for simulating motion and dynamics.

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Homework Statement



Why are (2) and (4) equations more accurately than (1) and (3) ? Why is 7*dt in (4) equation ? What kind of equations are (2) and (4) ? What method they used to write (3) and (4) equations?

Homework Equations



Velocity:
(1) v = (x - x[i-1]) / (t - t[i-1])
(2) v = (x[i+1] - x[i-1]) / (2*dt)

Acceleration:
(3) a = (v - v[i-1]) / (t - t[i-1])
(4) a = (2*x[i+2] - x[i+1] - 2*x - x[i-1] + 2*x[i-2]) / (7*dt)

The Attempt at a Solution


Verlet algorithm
Finite difference
 
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The problem with (1) and (3) is that there's a time shift between the input positions and the computed velocities and accelerations. E.g (1) is really estimating the velocity at step i-0.5. (2) corrects that.
(4) achieves the same result, but I'm not sure why it's quite as it is. (Shouldn't it have dt2 at the end?) If you start with the 'smoothed' acceleration expression (2*ai+1+3*ai+2*ai-1)/7 and then substitute for those ai using the forms ai.Δt = vi+.5-vi-.5 and vi.Δt = xi+.5-xi-.5 you arrive at (4) (with dt2). But why start with (2*ai+1+3*ai+2*ai-1)/7 rather than e.g. (ai+1+2*ai+ai-1)/4 I don't know.
 

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